加到 X 上的最小值，至少是 N 的 Y %
给定三个整数 N 、 X 和 Y ，任务是找出应该加到 X 上的最小整数，使其至少为 N 的 Y 。
示例:
输入: N = 10，X = 2，Y = 40 输出:2 X 加 2 得到 4，4 是 10 的 40%
输入: N = 10，X = 2，Y = 20 输出: 0 X 已经是 10 的 20%
方法:找到值= (N * Y) / 100 ，这是 N 的 Y 百分比。现在为了使 X 等于 val ，val–X必须加到 X 上，前提是 X < val 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the required value
// that must be added to x so that
// it is at least y percent of n
int minValue(int n, int x, int y)
{
// Required value
float val = (y * n) / 100;
// If x is already >= y percent of n
if (x >= val)
return 0;
else
return (ceil(val) - x);
}
// Driver code
int main()
{
int n = 10, x = 2, y = 40;
cout << minValue(n, x, y);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.lang.Math;
class GFG
{
// Function to return the required value
// that must be added to x so that
// it is at least y percent of n
static int minValue(int n, int x, int y)
{
// Required value
float val = (y * n) / 100;
// If x is already >= y percent of n
if (x >= val)
return 0;
else
return (int)(Math.ceil(val)-x);
}
// Driver code
public static void main(String[] args)
{
int n = 10, x = 2, y = 40;
System.out.println(minValue(n, x, y));
}
}
// This code is contributed by Code_Mech.
Python 3
import math
# Function to return the required value
# that must be added to x so that
# it is at least y percent of n
def minValue(n, x, y):
# Required value
val = (y * n)/100
# If x is already >= y percent of n
if x >= val:
return 0
else:
return math.ceil(val) - x
# Driver code
n = 10; x = 2; y = 40
print(minValue(n, x, y))
# This code is contributed by Shrikant13
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the required value
// that must be added to x so that
// it is at least y percent of n
static int minValue(int n, int x, int y)
{
// Required value
float val = (y * n) / 100;
// If x is already >= y percent of n
if (x >= val)
return 0;
else
return (int)(Math.Ceiling(val)-x) ;
}
// Driver code
public static void Main()
{
int n = 10, x = 2, y = 40;
Console.WriteLine((int)minValue(n, x, y));
}
}
// This code is contributed by Ryuga.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// php implementation of the approach
// Function to return the required value
// that must be added to x so that
// it is at least y percent of n
function minValue($n, $x, $y)
{
// Required value
$val = ($y * $n) / 100;
// If x is already >= y percent of n
if ($x >= $val)
return 0;
else
return (ceil($val) - $x);
}
// Driver code
{
$n = 10; $x = 2; $y = 40;
echo(minValue($n, $x, $y));
}
// This code is contributed by Code_Mech.
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the required value
// that must be added to x so that
// it is at least y percent of n
function minValue(n, x, y)
{
// Required value
let val = (y * n) / 100;
// If x is already >= y percent of n
if (x >= val)
return 0;
else
return (Math.ceil(val) - x);
}
// Driver code
let n = 10, x = 2, y = 40;
document.write(minValue(n, x, y));
// This code is contributed by souravmahato348
</script>
Output:
2
时间复杂度: O(1)
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